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Home 2019 June SQL Puzzle: SQL Advance Query – Place NULL for repeating values

SQL Puzzle: SQL Advance Query – Place NULL for repeating values

This article is half-done without your Comment! *** Please share your thoughts via Comment ***

Check the below input data and expected output to place NULL for repeating values.
Input Data:

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A B C Code
----------- ----------- ----------- ----
1 1 1 A
1 1 2 A
1 1 3 A
1 1 4 A
2 2 1 C
2 2 2 C
2 2 3 C

Expected Output:

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A B C Code
----------- ----------- ----------- ----
1 1 1 A
NULL NULL 2 NULL
NULL NULL 3 NULL
NULL NULL 4 NULL
2 2 1 C
NULL NULL 2 NULL
NULL NULL 3 NULL

Create a table with sample data:

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CREATE TABLE Code
(
A int
,B Int
,C int
,Code CHAR(1)
)
GO
INSERT INTO Code
VALUES
(1,1,1,'A')
,(1,1,2,'A')
,(1,1,3,'A')
,(1,1,4,'A')
,(2,2,1,'C')
,(2,2,2,'C')
,(2,2,3,'C')
GO

Solution:

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;with ctetest as
(
select *, row_number()over (partition by Code,A,B order by A) as rnk from Code
)
select
case when rnk = 1 then A else NULL end as A
,case when rnk = 1 then B else NULL end AS B
,C
,case when rnk = 1 then Code else NULL end AS Code
from ctetest

Please try the different solutions for this puzzle and share it via comment...

Jun 3, 2019Anvesh Patel
SQL Server: Script to find the total size of the IndexesPostgreSQL: Disable Non-Durable parameters and Improve Server Performance
Comments: 4
  1. Nilesh
    July 6, 2019 at 10:48 am

    Select
    Case When C = 1 then A Else NULL END as A,
    Case When C = 1 then B Else NULL END as B,
    C,
    Case When C = 1 then Code Else NULL END as Code
    From Code

    ReplyCancel
  2. Nikhil Bhosale
    August 13, 2019 at 5:26 am

    Select
    (Case when C != ‘1’ then NULL else A end) AS A,
    (Case when C != ‘1’ then NULL else B end) AS B,
    C,
    (Case when C != ‘1’ then NULL else code end) AS code
    from code

    ReplyCancel
  3. Phanindra Suripeddi
    October 9, 2019 at 6:15 pm

    DECLARE @Code TABLE
    (
    A int
    ,B Int
    ,C int
    ,Code CHAR(1)
    )

    INSERT INTO @Code
    VALUES
    (1,1,1,’A’)
    ,(1,1,2,’A’)
    ,(1,1,3,’A’)
    ,(1,1,4,’A’)
    ,(2,2,1,’C’)
    ,(2,2,2,’C’)
    ,(2,2,3,’C’)

    ;With T As
    (
    select *, checksum(concat(A,B,Code)) chksum From @code
    )
    ,T2 as
    (
    select *,lag(chksum) Over(partition by code Order by(select null)) as lag_chksum From T
    ) select
    case When chksum = lag_chksum then NULL else A End as A
    ,case When chksum = lag_chksum then NULL else B End as B
    ,C
    ,case When chksum = lag_chksum then NULL else code End as code
    From T2

    ReplyCancel
  4. koustav
    October 30, 2019 at 9:40 am

    select
    case when rnk = 1 then A else NULL end as A
    ,case when rnk = 1 then B else NULL end AS B
    ,C
    ,case when rnk = 1 then Code else NULL end AS Code from
    (
    select *, row_number()over (partition by Code,A,B order by A) as rnk from Code
    ) T1

    ReplyCancel

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Anvesh Patel
Anvesh Patel

Database Engineer

June 3, 2019 4 Comments SQL PuzzleAnvesh Patel, database, database research and development, dbrnd, Repeat, SQL Advance Query, SQL Interview, SQL Problem, SQL Programming, SQL Puzzle, SQL Query, SQL Tips and Tricks
About Me!

I'm Anvesh Patel, a Database Engineer certified by Oracle and IBM. I'm working as a Database Architect, Database Optimizer, Database Administrator, Database Developer. Providing the best articles and solutions for different problems in the best manner through my blogs is my passion. I have more than six years of experience with various RDBMS products like MSSQL Server, PostgreSQL, MySQL, Greenplum and currently learning and doing research on BIGData and NoSQL technology. -- Hyderabad, India.

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